\(\int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx\) [367]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [B] (verification not implemented)
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 27, antiderivative size = 103 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a x}{16}-\frac {a \cos ^5(c+d x)}{5 d}+\frac {a \cos ^7(c+d x)}{7 d}+\frac {a \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac {a \cos ^5(c+d x) \sin (c+d x)}{6 d} \] Output:

1/16*a*x-1/5*a*cos(d*x+c)^5/d+1/7*a*cos(d*x+c)^7/d+1/16*a*cos(d*x+c)*sin(d 
*x+c)/d+1/24*a*cos(d*x+c)^3*sin(d*x+c)/d-1/6*a*cos(d*x+c)^5*sin(d*x+c)/d
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.79 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a (420 d x-315 \cos (c+d x)-105 \cos (3 (c+d x))+21 \cos (5 (c+d x))+15 \cos (7 (c+d x))+105 \sin (2 (c+d x))-105 \sin (4 (c+d x))-35 \sin (6 (c+d x)))}{6720 d} \] Input:

Integrate[Cos[c + d*x]^4*Sin[c + d*x]^2*(a + a*Sin[c + d*x]),x]
 

Output:

(a*(420*d*x - 315*Cos[c + d*x] - 105*Cos[3*(c + d*x)] + 21*Cos[5*(c + d*x) 
] + 15*Cos[7*(c + d*x)] + 105*Sin[2*(c + d*x)] - 105*Sin[4*(c + d*x)] - 35 
*Sin[6*(c + d*x)]))/(6720*d)
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.08, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3042, 3317, 3042, 3045, 244, 2009, 3048, 3042, 3115, 3042, 3115, 24}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \sin ^2(c+d x) \cos ^4(c+d x) (a \sin (c+d x)+a) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (c+d x)^2 \cos (c+d x)^4 (a \sin (c+d x)+a)dx\)

\(\Big \downarrow \) 3317

\(\displaystyle a \int \cos ^4(c+d x) \sin ^3(c+d x)dx+a \int \cos ^4(c+d x) \sin ^2(c+d x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a \int \cos (c+d x)^4 \sin (c+d x)^2dx+a \int \cos (c+d x)^4 \sin (c+d x)^3dx\)

\(\Big \downarrow \) 3045

\(\displaystyle a \int \cos (c+d x)^4 \sin (c+d x)^2dx-\frac {a \int \cos ^4(c+d x) \left (1-\cos ^2(c+d x)\right )d\cos (c+d x)}{d}\)

\(\Big \downarrow \) 244

\(\displaystyle a \int \cos (c+d x)^4 \sin (c+d x)^2dx-\frac {a \int \left (\cos ^4(c+d x)-\cos ^6(c+d x)\right )d\cos (c+d x)}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle a \int \cos (c+d x)^4 \sin (c+d x)^2dx-\frac {a \left (\frac {1}{5} \cos ^5(c+d x)-\frac {1}{7} \cos ^7(c+d x)\right )}{d}\)

\(\Big \downarrow \) 3048

\(\displaystyle a \left (\frac {1}{6} \int \cos ^4(c+d x)dx-\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {a \left (\frac {1}{5} \cos ^5(c+d x)-\frac {1}{7} \cos ^7(c+d x)\right )}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle a \left (\frac {1}{6} \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx-\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {a \left (\frac {1}{5} \cos ^5(c+d x)-\frac {1}{7} \cos ^7(c+d x)\right )}{d}\)

\(\Big \downarrow \) 3115

\(\displaystyle a \left (\frac {1}{6} \left (\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {a \left (\frac {1}{5} \cos ^5(c+d x)-\frac {1}{7} \cos ^7(c+d x)\right )}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle a \left (\frac {1}{6} \left (\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {a \left (\frac {1}{5} \cos ^5(c+d x)-\frac {1}{7} \cos ^7(c+d x)\right )}{d}\)

\(\Big \downarrow \) 3115

\(\displaystyle a \left (\frac {1}{6} \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {a \left (\frac {1}{5} \cos ^5(c+d x)-\frac {1}{7} \cos ^7(c+d x)\right )}{d}\)

\(\Big \downarrow \) 24

\(\displaystyle a \left (\frac {1}{6} \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )-\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {a \left (\frac {1}{5} \cos ^5(c+d x)-\frac {1}{7} \cos ^7(c+d x)\right )}{d}\)

Input:

Int[Cos[c + d*x]^4*Sin[c + d*x]^2*(a + a*Sin[c + d*x]),x]
 

Output:

-((a*(Cos[c + d*x]^5/5 - Cos[c + d*x]^7/7))/d) + a*(-1/6*(Cos[c + d*x]^5*S 
in[c + d*x])/d + ((Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (3*(x/2 + (Cos[c + 
 d*x]*Sin[c + d*x])/(2*d)))/4)/6)
 

Defintions of rubi rules used

rule 24
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 244
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand 
Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p 
, 0]
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3045
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_ 
Symbol] :> Simp[-(a*f)^(-1)   Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], 
x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && 
 !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])
 

rule 3048
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_), x_Symbol] :> Simp[(-a)*(b*Cos[e + f*x])^(n + 1)*((a*Sin[e + f*x])^(m - 
1)/(b*f*(m + n))), x] + Simp[a^2*((m - 1)/(m + n))   Int[(b*Cos[e + f*x])^n 
*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] 
 && NeQ[m + n, 0] && IntegersQ[2*m, 2*n]
 

rule 3115
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

rule 3317
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a   Int[(g*Co 
s[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Simp[b/d   Int[(g*Cos[e + f*x])^ 
p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]
 
Maple [A] (verified)

Time = 16.89 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.81

method result size
parallelrisch \(-\frac {\left (-4 d x +\sin \left (4 d x +4 c \right )+\frac {\sin \left (6 d x +6 c \right )}{3}+3 \cos \left (d x +c \right )+\cos \left (3 d x +3 c \right )-\frac {\cos \left (5 d x +5 c \right )}{5}-\frac {\cos \left (7 d x +7 c \right )}{7}-\sin \left (2 d x +2 c \right )+\frac {128}{35}\right ) a}{64 d}\) \(83\)
derivativedivides \(\frac {a \left (-\frac {\sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{5}}{7}-\frac {2 \cos \left (d x +c \right )^{5}}{35}\right )+a \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{5}}{6}+\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )}{d}\) \(88\)
default \(\frac {a \left (-\frac {\sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{5}}{7}-\frac {2 \cos \left (d x +c \right )^{5}}{35}\right )+a \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{5}}{6}+\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )}{d}\) \(88\)
risch \(\frac {a x}{16}-\frac {3 a \cos \left (d x +c \right )}{64 d}+\frac {a \cos \left (7 d x +7 c \right )}{448 d}-\frac {a \sin \left (6 d x +6 c \right )}{192 d}+\frac {a \cos \left (5 d x +5 c \right )}{320 d}-\frac {a \sin \left (4 d x +4 c \right )}{64 d}-\frac {a \cos \left (3 d x +3 c \right )}{64 d}+\frac {a \sin \left (2 d x +2 c \right )}{64 d}\) \(108\)
norman \(\frac {\frac {4 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}+\frac {a x}{16}-\frac {4 a}{35 d}-\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}+\frac {11 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}-\frac {31 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{24 d}+\frac {31 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{24 d}-\frac {11 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{6 d}+\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{8 d}+\frac {7 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{16}+\frac {21 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{16}+\frac {35 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{16}+\frac {35 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{16}+\frac {21 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{16}+\frac {7 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{16}+\frac {a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{16}-\frac {4 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{5 d}-\frac {8 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}-\frac {4 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}+\frac {8 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{5 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{7}}\) \(318\)
orering \(\text {Expression too large to display}\) \(2407\)

Input:

int(cos(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

-1/64*(-4*d*x+sin(4*d*x+4*c)+1/3*sin(6*d*x+6*c)+3*cos(d*x+c)+cos(3*d*x+3*c 
)-1/5*cos(5*d*x+5*c)-1/7*cos(7*d*x+7*c)-sin(2*d*x+2*c)+128/35)*a/d
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.71 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {240 \, a \cos \left (d x + c\right )^{7} - 336 \, a \cos \left (d x + c\right )^{5} + 105 \, a d x - 35 \, {\left (8 \, a \cos \left (d x + c\right )^{5} - 2 \, a \cos \left (d x + c\right )^{3} - 3 \, a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{1680 \, d} \] Input:

integrate(cos(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="fricas" 
)
 

Output:

1/1680*(240*a*cos(d*x + c)^7 - 336*a*cos(d*x + c)^5 + 105*a*d*x - 35*(8*a* 
cos(d*x + c)^5 - 2*a*cos(d*x + c)^3 - 3*a*cos(d*x + c))*sin(d*x + c))/d
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 192 vs. \(2 (90) = 180\).

Time = 0.49 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.86 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\begin {cases} \frac {a x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {3 a x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {3 a x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {a x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {a \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {a \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} - \frac {a \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac {a \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac {2 a \cos ^{7}{\left (c + d x \right )}}{35 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right ) \sin ^{2}{\left (c \right )} \cos ^{4}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:

integrate(cos(d*x+c)**4*sin(d*x+c)**2*(a+a*sin(d*x+c)),x)
 

Output:

Piecewise((a*x*sin(c + d*x)**6/16 + 3*a*x*sin(c + d*x)**4*cos(c + d*x)**2/ 
16 + 3*a*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + a*x*cos(c + d*x)**6/16 + a 
*sin(c + d*x)**5*cos(c + d*x)/(16*d) + a*sin(c + d*x)**3*cos(c + d*x)**3/( 
6*d) - a*sin(c + d*x)**2*cos(c + d*x)**5/(5*d) - a*sin(c + d*x)*cos(c + d* 
x)**5/(16*d) - 2*a*cos(c + d*x)**7/(35*d), Ne(d, 0)), (x*(a*sin(c) + a)*si 
n(c)**2*cos(c)**4, True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.63 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {192 \, {\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} a + 35 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a}{6720 \, d} \] Input:

integrate(cos(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="maxima" 
)
 

Output:

1/6720*(192*(5*cos(d*x + c)^7 - 7*cos(d*x + c)^5)*a + 35*(4*sin(2*d*x + 2* 
c)^3 + 12*d*x + 12*c - 3*sin(4*d*x + 4*c))*a)/d
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.04 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {1}{16} \, a x + \frac {a \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac {a \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac {a \cos \left (3 \, d x + 3 \, c\right )}{64 \, d} - \frac {3 \, a \cos \left (d x + c\right )}{64 \, d} - \frac {a \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {a \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {a \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \] Input:

integrate(cos(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="giac")
 

Output:

1/16*a*x + 1/448*a*cos(7*d*x + 7*c)/d + 1/320*a*cos(5*d*x + 5*c)/d - 1/64* 
a*cos(3*d*x + 3*c)/d - 3/64*a*cos(d*x + c)/d - 1/192*a*sin(6*d*x + 6*c)/d 
- 1/64*a*sin(4*d*x + 4*c)/d + 1/64*a*sin(2*d*x + 2*c)/d
 

Mupad [B] (verification not implemented)

Time = 20.20 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.83 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a\,x}{16}+\frac {\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{8}-\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{6}+\left (\frac {a\,\left (2205\,c+2205\,d\,x-6720\right )}{1680}-\frac {21\,a\,\left (c+d\,x\right )}{16}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\frac {31\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{24}+\left (\frac {a\,\left (3675\,c+3675\,d\,x+6720\right )}{1680}-\frac {35\,a\,\left (c+d\,x\right )}{16}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\left (\frac {a\,\left (3675\,c+3675\,d\,x-13440\right )}{1680}-\frac {35\,a\,\left (c+d\,x\right )}{16}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-\frac {31\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{24}+\left (\frac {a\,\left (2205\,c+2205\,d\,x+2688\right )}{1680}-\frac {21\,a\,\left (c+d\,x\right )}{16}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6}+\left (\frac {a\,\left (735\,c+735\,d\,x-1344\right )}{1680}-\frac {7\,a\,\left (c+d\,x\right )}{16}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}+\frac {a\,\left (105\,c+105\,d\,x-192\right )}{1680}-\frac {a\,\left (c+d\,x\right )}{16}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^7} \] Input:

int(cos(c + d*x)^4*sin(c + d*x)^2*(a + a*sin(c + d*x)),x)
 

Output:

(a*x)/16 + ((a*(105*c + 105*d*x - 192))/1680 - (a*tan(c/2 + (d*x)/2))/8 - 
(a*(c + d*x))/16 + tan(c/2 + (d*x)/2)^2*((a*(735*c + 735*d*x - 1344))/1680 
 - (7*a*(c + d*x))/16) + tan(c/2 + (d*x)/2)^4*((a*(2205*c + 2205*d*x + 268 
8))/1680 - (21*a*(c + d*x))/16) + tan(c/2 + (d*x)/2)^10*((a*(2205*c + 2205 
*d*x - 6720))/1680 - (21*a*(c + d*x))/16) + tan(c/2 + (d*x)/2)^8*((a*(3675 
*c + 3675*d*x + 6720))/1680 - (35*a*(c + d*x))/16) + tan(c/2 + (d*x)/2)^6* 
((a*(3675*c + 3675*d*x - 13440))/1680 - (35*a*(c + d*x))/16) + (11*a*tan(c 
/2 + (d*x)/2)^3)/6 - (31*a*tan(c/2 + (d*x)/2)^5)/24 + (31*a*tan(c/2 + (d*x 
)/2)^9)/24 - (11*a*tan(c/2 + (d*x)/2)^11)/6 + (a*tan(c/2 + (d*x)/2)^13)/8) 
/(d*(tan(c/2 + (d*x)/2)^2 + 1)^7)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.11 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \left (-240 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6}-280 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5}+384 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}+490 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}-48 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-105 \cos \left (d x +c \right ) \sin \left (d x +c \right )-96 \cos \left (d x +c \right )+105 d x +96\right )}{1680 d} \] Input:

int(cos(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c)),x)
 

Output:

(a*( - 240*cos(c + d*x)*sin(c + d*x)**6 - 280*cos(c + d*x)*sin(c + d*x)**5 
 + 384*cos(c + d*x)*sin(c + d*x)**4 + 490*cos(c + d*x)*sin(c + d*x)**3 - 4 
8*cos(c + d*x)*sin(c + d*x)**2 - 105*cos(c + d*x)*sin(c + d*x) - 96*cos(c 
+ d*x) + 105*d*x + 96))/(1680*d)